我在MYSQL中创建了表格,其中列有' name',' worker_id',' local_job_id'这里我的php代码给出了给定local_job_id的相应名称 来自链接
http://allwaysready.16mb.com/Cuboid/LocalCheckRequest.php?local_job_id[]=3
给出
{"result":[{"worker_id":"2"},{"worker_id":"1"}]}
但是,如果worker_id和local_job_id都匹配,我想在此代码中更改的内容是显示名称。
" http://allwaysready.16mb.com/Cuboid/LocalCheckRequest.php?local_job_id[]=3=2&worker_id=2"那个名字'应显示对应于local_job_id = 3和worker_id = 2的内容。
<?php
if ($_SERVER['REQUEST_METHOD'] == 'GET') {
$local_job_id = $_GET['local_job_id'];
require_once ('dbConnect.php');
$local_job_request = array();
foreach($_REQUEST['local_job_id'] as $key => $val) {
$local_job_request[$key] = filter_var($val, FILTER_SANITIZE_STRING);
}
$local_job_ids = "'" . implode("','", $local_job_request) . "'";
$sql = "SELECT * FROM local_job_request WHERE local_job_id IN ({$local_job_ids})";
$r = mysqli_query($con, $sql);
$result = array();
while ($res = mysqli_fetch_array($r)) {
array_push($result, array(
"worker_id" => $res['worker_id']
));
}
echo json_encode(array(
"result" => $result
));
mysqli_close($con);
}
答案 0 :(得分:1)
尝试这可能对您有所帮助
if($_SERVER['REQUEST_METHOD']=='GET'){
$local_job_id = $_REQUEST['local_job_id'];
$worker_id = $_GET['worker_id'];
require_once('dbConnect.php');
$local_job_request = array();
foreach ($local_job_id as $key => $val) {
$local_job_request[$key] = filter_var($val, FILTER_SANITIZE_STRING);
}
$local_job_ids = "'" . implode("','", $local_job_request) . "'";
$sql = "SELECT * FROM local_job_request WHERE local_job_id IN ({$local_job_ids})";
if(isset($worker_id) && $worker_id != "") {
$sql .= " AND worker_id = ".$worker_id;
}
$r = mysqli_query($con,$sql);
$result = array();
while($res = mysqli_fetch_array($r)){
array_push($result,array(
"worker_id"=>$res['worker_id']
)
);
}
echo json_encode(array("result"=>$result));
mysqli_close($con);
}